Figuring odds...

[Homerboy]

I was never too good at such things... but we happen to have 3 people here in the office with the same birthday (today).

What are the odds of this happening in an office of 75 people?

 

[Joemonkey]

apparently about 3 out of 75

😉

 

[Homerboy]

Originally posted by: Joemonkey
apparently about 3 out of 75

😉

don't you have emails to delete and tracks to cover?

 

[Joemonkey]

My contingency plan was complete after ~30 minutes... i'm good to nef for the next 2 weeks

anyway, click here

the birthday paradox thingy is a popular math problem

"With two people, there are 363 unused birthdays. The probability that a third person has a birthday that differs from the other two distinct birthdays is 363/365. So, for three people, the chance of having no pair of matching birthdays is 365/365 x 364/365 x 363/365, or .9918.

As the number of people brought into the group increases, the chance of there being no match decreases. By the time the crowd numbers 23 people, the probability of no matching birthdays is .4927. Thus, the chance of at least one match within a group of 23 people is .5073, or slightly better than 50 percent. "

it doesn't get into 3 people sharing a birthday, but i'm pretty sure the math can be continued out that way

 

[dullard]

Originally posted by: Joemonkey
it doesn't get into 3 people sharing a birthday, but i'm pretty sure the math can be continued out that way

It does too get into 3 people sharing a birthday. You just didn't read far enough. If people were randomly born on each day (almost true) then with 88 people, you'd have a 50% chance of 3 people sharing a birthday. Since Homerboy has 75 people, his office is a bit under 50%.

 

[Homerboy]

50%? That seems like pretty good odds.